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3 years ago

13

Brandon used inverse operations to solve an equation in his homework assignment. During which step, if any at all, did Brandon u

se the incorrect inverse operation to isolate the variable?
Equation: 8x3 + 2 = 26

Brandon's solution:

Step 1
Step 2
All of the inverse operations were correct.
Step 3

2 answers:

ICE Princess25 [194]3 years ago

8 0

Answer:

Step 3

Step-by-step explanation:

x^3=3

you need to root cube both side not divide both side by 3

You cannot just divide the cube (^3) by 3 to make it 1/

Evgesh-ka [11]3 years ago

3 0

ANSWER: He did it right

Step-by-step explanation:

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Ratios equivalent to 2:6.

Alexandra [31]

8:24
4:12
100:300

All you have to do is multiply both sides by the same number to get an equivalent ratio.

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This diagram of a rectangular city park was drawn using a scale of 1 centimeter to 20 meters

I am Lyosha [343]

Answer:

Scale factor for the drawing from actual park = $\frac{1}{20}$

Area of the actual park = 1200 cm²

Step-by-step explanation:

Length of the rectangular city park = 5 cm

Width of the rectangular park = 6 cm

Using scale factor 1 cm = 20 meters

Scale factor = $\frac{\text{length of the park in drawing}}{\text{Actual length of park}}=\frac{1}{20}$

$\frac{5}{\text{Actual length of park}}=\frac{1}{20}$

Actual length = 5 × 20 = 100 meters

Actual width = 6 × 20 = 120 meters

Area of the rectangular park = length × width

= 100 × 120

= 1200 square meters

Therefore, Scale factor from actual length to the length in drawing = 1 : 20

Area of the rectangular park = 1200 square feet

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3 years ago

a high school student works two summer jobs . One at a coffee shop and one at a grocery store. The coffee shop pays 10$ an hour

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That's $600....so....he needs 200 more
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3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago